The claim is proved. Therefore, \((ii)\)  holds, and \(l\ge (k_1+k_2)\mathcal {S}^2\) . Moreover, by [1]}, \(m^-(a_1,0)\rightarrow \frac{\mu _1}{4}\mathcal {S}^2\)  as \(a_1\rightarrow 0\) , and \(m^-(0,a_2)\rightarrow \frac{\mu _2}{4}\mathcal {S}^2\)  as \(a_2\rightarrow 0\) . We obtain that
\(\begin{aligned}\frac{k_1+k_2}{4}\mathcal {S}^2+m^+(a_1,a_2)&>m^-(a_1,a_2)=I(u^-_{a_1},v^-_{a_2})\\&=\frac{1}{4}\int _{\mathbb {R}^4}|\nabla u^-_{a_1}|^2+|\nabla v^-_{a_2}|^2-\frac{\alpha _1}{p}\big (1-\frac{p\gamma _p}{4}\big )\int _{\mathbb {R}^4}|u^-_{a_1}|^p\\&\quad -\frac{\alpha _2}{p}\big (1-\frac{p\gamma _p}{4}\big )\int _{\mathbb {R}^4}|v^-_{a_2}|^p\\&\ge \frac{1}{4}\int _{\mathbb {R}^4}|\nabla u^-_{a_1}|^2+|\nabla v^-_{a_2}|^2+o(1)\\&\ge \frac{k_1+k_2}{4}\mathcal {S}^2+o(1).\end{aligned}\) 
